Question

Find the point on the $x-$axis that is equidistant from $(1,2)$ and $(3,-4)$.

Answer 1

Step 1: Determine distance from each point

Let $A(1,2)$ and $B(3,-4)$ be the two points that are equidistant from $P(x,0)$.

Here we consider $y=0$ because the point lies on x-axis.
Using distance formula,

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$PA=\sqrt{{\left(x-1\right)}^2+{\left(0-2\right)}^2}$

$PB=\sqrt{{\left(x-3\right)}^2+{\left(0+4\right)}^2}$

Step 2: Determine $x$

It is given that $PA=PB$

$\therefore \sqrt{{\left(x-1\right)}^2+{\left(0-2\right)}^2}=\sqrt{{\left(x-3\right)}^2+{\left(0+4\right)}^2}$

$\Rightarrow$. ${\left(x-1\right)}^2+{\left(0-2\right)}^2={\left(x-3\right)}^2+{\left(0+4\right)}^2$
$\Rightarrow$$x^2+1–2x+4=x^2+9–6x+16$
$\Rightarrow$ $-2x+5=-6x+25$
$\Rightarrow$ $4x=20$
$\Rightarrow$ $x=5$
Hence, the point on the $x-$axis that is equidistant from $(1,2)$ and $(3,-4)$ is $(5,0)$.